(II.G) PRIME POWER MODULI AND POWER RESIDUES

Transcription

1 II.G PRIME POWER MODULI AND POWER RESIDUES I II.C, we used the Chiese Remider Theorem to reduce cogrueces modulo m r i i to cogrueces modulo r i i. For exmles d roblems, we stuck with r i 1 becuse we hd o techiques to reduce from mod r i i to mod i. We ow discuss oe roch to this. 1 Let be rime, f Z[x] olyomil. Theorem 1 Hesel s Lemm. Let α Z/ j Z be solutio of f x j 0. The there exists uique solutio α Z/ j+1 Z of 0, with f α / f x j+1 0, with α j α. We sy tht α lifts α. Proof. Let be iteger reducig to α mod j, so tht f j Writig deg f, cosider the Tylor exsio f + t j f + t j f + t j f! + + t j f,! where the f k k! re itegers. Why? Reducig mod j+1, we hve f + t j d so t solves f + t j j+1 0 j+1 f + t j f ; t solves t j f f j+1 f 0 j f / 0 t solves t f f j t f j f. 1 If you wt to kow more, see [NZM].6. Hit: the m k mm 1 m k+1 k! re itegers for ech m k. 1 0.

2 MATH II.G This uiquely secifies the cogruece clss α of + t j modulo j+1. Exmle. Let m {1,,..., 1}. Whe does x m hve solutio? Tht is, whe does m hve squre root i Z/ Z? Clerly x x m, so m must be squre mod which is true for hlf of {1,,..., 1}. Suose this is true: m 1, i.e. f 1 : 1 m 0. The we hve the bse cse for the followig iductio: if f x 0 hs solutio j liftig 1, the j f j j / 0 sice j 1 / 0, d / 0, so Hesel imlies the existece of j+1 liftig j hece 1 such tht f j+1 j+1 0. Coclude tht m exists i Z/ Z m exists i Z/Z. For istce, if m d 7, or 4, sy 3. Now to costruct squre root mod 7 we set t }{{} t 1 4t : 6t t the to lift mod t }{{} t 94t : 3 6t 7 t 3 108

3 MATH II.G 3 d so o. So you see tht i this cse cogruece roblem for got reduced to oe mod. We ow use Hesel s Lemm to del with rimitive roots. Here will be odd rime. Ste 1 mod : Let 3 g be rimitive root mod, d t Z/Z. The g + t, g, 1 g + t, 1 g + t belogs to Z/ Z. Deote its order by h, d ote tht h Z/ Z φ 1. Moreover, g + t h 1 g h g + t h 1, which by Fermt d the fct tht g geertes Z/Z gives 1 h. So h 1 or 1. Now set f x : x 1 1, α g, d ote f g 1g / 0. By Hesel s Lemm, there is uique solutio of f x 0, i.e. uique choice of t so tht h 1. So for other choices, h 1 d g + t is rimitive root mod. Ste mod r> : Let g be rimitive root mod ; deote the order of g mod r by h r, which must stisfy h r φ r r 1 1 by Lgrge. Moreover, g h r 1 r gh r 1 1 h r sice g is rimitive mod h r β 1 with β {1,..., r 1}. We clim tht β r 1, i.e. g r 1 / 1 for ech r r. Sice we kow this for r, ssume it iductively for,..., r d rove for r + 1. Tht is, by ssumtio, g r 1 is but / 1, so tht we my write r g r t r 1, t. r More recisely: g is some iteger whose cogruece clss mod geertes Z/Z. Sme thig t the begiig of Ste with relcig.

4 4 MATH II.G Tkig th owers gives g r t r t r { } t r 1 +, d sice for k r+1 k kr 1 the brcketed terms die mod r+1, levig us with g r 1 1 This roves the existece rt of 1 + t r / 1. r+1 r+1 Theorem 3. There exist r 1φ 1 rimitive roots mod r y odd rime, r 1. I rticulr, Z/ r Z is cyclic. Proof. Existece of sigle rimitive root doe bove roves cyclicity, i.e. tht Z/ r Z, Z/φ r Z, +. Recllig tht the order of m i Z/Z, + is, so tht m is geertor iff m, 1,,m we see tht there re φφ r φ r 1 1 φ r 1 φ 1 r 1φ 1 geertors. Power Residues. Let, s bove, be odd rime. Defiitio 4. A th ower residue mod is y Z/Z tht c be writte s th ower mod. For, this is clled qudrtic residue; if 3 cubic residue, d so o. Theorem 5. Z/Z is th ower residue 1, 1 1. I this cse, x hs, 1 solutios. Proof. Use the isomorhism Z/Z, Z/ 1Z, +, m g m

5 MATH II.G 5 where g is geertor of Z/Z. We hve g m for some m, so x for some x g y, sy m 1 y for some y Z/ 1Z, 1 m by Theorem II.B.5i 1 m 1, 1 1, 1 g m 1, 1 1. The umber of solutios comes directly from Theorem II.B.5ii. Sice 1,, we hve the Corollry 6 Euler s criterio. Z/Z is qudrtic residue We will use this i our discussio of qudrtic recirocity. By Theorem 5, every th ower residue, i.e. ech elemet i the imge of Z/Z Z/Z hs reimge cosistig of, 1 elemets. So the m is, 1 : 1, d the imge cotis Corollry 7. Exctly Exmle 8. Z/7Z hs 1, 1 elemets. 1, 1 elemets of Z/Z re th ower residues. 3 6 qudrtic residues: 1,, d 4. By Euler, these must,6 cube to 1. 6 cubic residues: 1 d 6. By Theorem 5, these must 3,6 squre to qurtic residues: sme s qudrtic. 4,6 6 6 quitic residues: everythig. 5,6 4 The oly other ossibility is to hve 1 1.

6 6 MATH II.G Qudrtic recirocity. We coclude our discussio of cogrueces with fmous d owerful result of Guss tht ids i determiig whe oe rime is qudrtic residue modulo other. Write QR for qudrtic residue d NR for qudrtic o-residue. Recll tht for odd rime, d iteger corime to, we hve Euler s criterio Corollry 6: Notice tht i fct is QR mod Z/Z 1 {+1, 1} Z/Z is grou homomorhism, which imlies the Corollry 9. QR QR QR, QR NR NR, d NR NR QR. You ll lso recll tht x 1 is solvble if d oly if 4 1 for odd rime; cf. Cor. II.B.9. We get very quick re-roof of tht result by Euler s criterio: Corollry is QR mod 1 4 Thm. II.B.10 Q 1. slits i Proof. 1 is eve iff 4 1. Aly Euler. Defiitio 11. For odd rime, the Legedre symbol is 0 if, : 1 if is QR mod, 1 if is NR mod. Some esy roerties of the symbol re: b b [by Cor. 9] +k 1 if, 1.

7 Exmle MATH II.G Lemm 13 Guss. Let be odd rime,, 1. Cosider the set S : {,, 3,..., 1 }, d reduce it modulo to subset S of { +1,..., 1, 0, 1,..., 1 }. Deote by ν the umber of egtive itegers i this subset; the 1 ν. Proof. Sice there is o ir of elemets of S with sum divisible by, o ir of the form {b, b} c belog to S. As S hece S cosists of 1 distict elemets, S cotis either 1 or 1, either or, d so o u to ± 1. Accordigly, write where ech ɛ i is +1 or 1. S {ɛ 1 1, ɛ,..., ɛ 1 1 } Multilyig ll the elemets of S together, d usig the fct tht S S, we hve 1 1 ɛ 1 1 1ɛ ɛ 1 which fter ccelltios becomes 1 ɛ 1 ɛ ɛ 1 1 ν, which by Euler s criterio is. Exmle 14 11, 3. S {3, 6, 9, 1, 15}, d S {3, 5,, 1, 4} hs two egtive umbers; so d 3 is QR mod 11 ideed Lemm 15. Let be odd rime, > 0 with, 1. The deeds oly o modulo 4: for rime q ±, q. 4 We will defer the roof to fter tht of the followig mi

8 8 MATH II.G Theorem 16 Qudrtic Recirocity Lw. Let d q be distict odd rimes. The q q 1 1 q 1. More cocretely, this result which we ll bbrevite QRL sys tht q q if 1 or q d q q if 3 q. 4 4 Proof of QRL. Cosider first the cse 4 q. We my ssume > q, so tht q + 4, > 0. The q q q q q while q which by Lemm 15 is 1 1 q, q 1 1 q. So the result follows i this cse. Now suose / q. The q, d so q + 4, > 0, 4 4 d q q q q q where we used Lemm 15 i the middle equlity. Remrk 17. How to del with the rime? Oe hs the followig comlemet to the QRL: 5 +1 if ± if ± A roof my be foud i [NZM] 3.3.

9 MATH II.G 9 Exmle 18. Sice 5 4 1, QRL So x 5 hs o solutios Similrly, sice 1, QRL We ow tur to the Proof of Lemm 15. As bove write S {,,..., 1 }. Put I, 3, b 1, b 7 where b. I clim tht every elemet of S which is to somethig i, 0, lies i I. First cosider the cse b. We hve b > 1, so the Clim is OK here. The other cse is where b 1 : from b > 1, it follows tht b 1, b is the lst itervl tht could coti elemet of S tht reduces to, 0. Clim is roved. With ν s i Lemm 13 d its roof, we hve 1 ν which by our Clim 1 S I. Now writig 1 1 S {1,, 3,..., } d 1 I, 3, b 1, b 0,, we fid S I 1 S 1 I Z 1 I. By the divisio lgorithm, we c write 4c + r, d ote tht J : r, r 3r, r b 1 r, br is just 1 I with edoits of itervls moved by eve itegers. So ν Z 1 I Z J, 6 Or, if you refer ot to use QRL, you c just comute 5 51 mod 103. Good luck with tht.

10 10 MATH II.G which lredy roves tht deedso oly modulo 4. We oly eed to check ow tht if q, we get the sme result. 4 I the bove comuttio, this mes relcig r by 4 r, hece J by r, 4 r 6 3r, 8 r 4b b 1 r, 4b br which is J with edoits moved by eve itegers. I rest my cse. There is geerliztio of the Legedre symbol to some comosite moduli. Defiitio 19. Let be ositive odd iteger, with rime fctoriztio α 1 1 α r r. The Jcobi symbol is give by or by 1, if 1. : k i1 The three esy roerties of the Legedre symbol, d the QRL, crry over verbtim. For istce, s log s d re odd d corime. The mi differece is tht 1 does ot imly tht is QR mod. This is becuse, by the Chiese Remider Theorem, Hesel QR mod while o the other hd CRT QR mod ech i 1 i αi QR mod ech α i i 1 i, k i1 i αi 1 is much weker. The Jcobi symbol lso leds to ew rimlity test: give, 1, clculte 1 d mod. If they differ, obviously is comosite. But wit: how do you comute without kowig α 1 1 α k k? i,

11 MATH II.G 11 By usig the QRL!! Fli it, fter reducig mod of course; lyig this reetedy is very similr to the Euclide Algorithm for comutig the GCD! Comuttiolly, this is the true imortce of qudrtic recirocity.